How much Hawking Radiation does an evaporating Black Hole emit in the last second?

[Yora]

Hawking radiation is sometimes thrown around as a massive source of power and the final moments of evaporation as a hugely catastrophic event.
Since black hole evaporation accelerates as the surface to volume ratio increases and we're dealing with a singularity, the acceleration will speed up dramatically towards the end. But the surface is also getting much smaller just as quickly and the region that emits Hawking radiation becomes tiny.

So, can we calculate how much energy is getting emmited during the last second? And milisecond, minute, hour, and day?

Would we see a massive explosion or perhaps nothing at all?

Also, half of the Hawking Radiation should be antimatter, so in the final moments you might also have additional matter-antimatter annihilations. Would that make things more dramatic?

[halfeye]

Hawking radiation is sometimes thrown around as a massive source of power and the final moments of evaporation as a hugely catastrophic event.
Since black hole evaporation accelerates as the surface to volume ratio increases and we're dealing with a singularity, the acceleration will speed up dramatically towards the end. But the surface is also getting much smaller just as quickly and the region that emits Hawking radiation becomes tiny.

So, can we calculate how much energy is getting emmited during the last second? And milisecond, minute, hour, and day?

Would we see a massive explosion or perhaps nothing at all?

Also, half of the Hawking Radiation should be antimatter, so in the final moments you might also have additional matter-antimatter annihilations. Would that make things more dramatic?

You don't need to know the exact amount of energy. What we know is the mass, and that E=Mc^2.

According to Jerry Pournelle in his book Black holes (copyright date 1978), a one second black hole has a mass of 2365 tons (presumably also pretty near that in tonnes) (though remember that the time is a half life, it's random in the way that half lives are). A nuclear bomb does E=Mc^2 to much less than one gram of matter. The final seconds of an evaporating black hole would be very energetic. They should also be almost exactly identical every time.

I think the anti-matter would not make a difference, it anihilates at E=Mc^2 so that's exactly the same amount of energy (in so far as any matter escapes far enough from the event horizon to not be immediately reabsorbed).

[Leewei]

Hawking radiation is sometimes thrown around as a massive source of power and the final moments of evaporation as a hugely catastrophic event.
Since black hole evaporation accelerates as the surface to volume ratio increases and we're dealing with a singularity, the acceleration will speed up dramatically towards the end. But the surface is also getting much smaller just as quickly and the region that emits Hawking radiation becomes tiny.

So, can we calculate how much energy is getting emmited during the last second? And milisecond, minute, hour, and day?

Would we see a massive explosion or perhaps nothing at all?

Also, half of the Hawking Radiation should be antimatter, so in the final moments you might also have additional matter-antimatter annihilations. Would that make things more dramatic?

Hawking visualized evaporation of a black hole as being much like its formation, only in reverse. This means that everything that ever fell into a black hole eventually comes back out as space un-warps itself around the vanishing singularity.

Hawking radiation would involve particles and anti-particles springing into existence at the Schwarzschild radius of the singularity. The anti-particles would annihilate some matter inside the singularity, while particles would accumulate there. The energy debt of creating these pairs is paid by the singularity, though, so those normally energetic matter/antimatter annihilations would not accrue any energy inside the black hole. What it would do is make the original matter of the singularity even less recognizable.

What you would see would depend on the original composition of the black hole. An evaporating hole would be somewhat isolated from matter and energy, or else it would feed and outgrow its decay. Therefore, the stuff it contained would probably be very hot, very dense matter. The singularity would be small enough that neutron matter would very quickly change into other types of matter. I expect the result would be a small superheated cloud of very radioactive plasma made of heavier elements.

[factotum]

Hawking visualized evaporation of a black hole as being much like its formation, only in reverse. This means that everything that ever fell into a black hole eventually comes back out as space un-warps itself around the vanishing singularity.

Um, I don't think that's how it works? The particle pairs which are the ultimate source of Hawking radiation are popping into existence randomly all over the universe, it's just that normally they annihilate each other so fast you don't notice. Near the black hole one of the pairs may fall into the black hole while the other escapes, though. Those particles are not particles that were originally absorbed by the black hole, although the energy that goes to make them comes from the hole.

[gomipile]

The particle pairs explanation is a flawed analogy, anyway. The calculations that result in Hawking radiation don't use pair production.

Look up the Unruh effect. The actual explanation for Hawking radiation is essentially the same, only using a spherically symmetric (black hole) spacetime instead of a flat spacetime which appears curved due to the constant acceleration of an observer.

[Lord Torath]

The particle pairs explanation is a flawed analogy, anyway. The calculations that result in Hawking radiation don't use pair production.

Look up the Unruh effect. The actual explanation for Hawking radiation is essentially the same, only using a spherically symmetric (black hole) spacetime instead of a flat spacetime which appears curved due to the constant acceleration of an observer.

Huh. I've only heard the particle/anti-particle pairs explanation. I'll have to look into this Unruh effect. Something new to learn! Yay!

[Leewei]

Um, I don't think that's how it works? The particle pairs which are the ultimate source of Hawking radiation are popping into existence randomly all over the universe, it's just that normally they annihilate each other so fast you don't notice. Near the black hole one of the pairs may fall into the black hole while the other escapes, though. Those particles are not particles that were originally absorbed by the black hole, although the energy that goes to make them comes from the hole.

The text you quoted isn't about Hawking radiation, but about what happens to the mass that originally formed the singularity. Since that mass would probably be neutronium (the last state matter transitions into prior to collapse into a singularity), the last stuff coming out of a black hole would be the same sort of stuff.

And recently, we've had a glimpse at what happens when neutronium gets dispersed, thanks in large part to LIGO.

[Yora]

Nothing comes out of a black hole, even when it evaporates. It's assumed to be a singularity of infinite density, so no matter how much it shrinks, it will not lose its event horizon and release anything.
Even Hawking radiation does not come from inside the black hole but from just outside of the event horizon.

[Leewei]

Nothing comes out of a black hole, even when it evaporates. It's assumed to be a singularity of infinite density, so no matter how much it shrinks, it will not lose its event horizon and release anything.
Even Hawking radiation does not come from inside the black hole but from just outside of the event horizon.

You're not down with Hawking's "Apparent Horizon" idea, I take it?

I'll be more specific. An event horizon has a very firm set of rules, one of which is that it's strictly a one-way item. The problem with this is that it violates conservation of information, which kind of breaks physics. The notion is something that troubled Hawking, so he mulled it over for a while, then announced that he had a solution. Gravity was not simply a constant effect, so event horizons are not quite the point (well, sphere) of no return we thought they were. Fluctuations of gravity could actually allow mass to escape from black holes. Due to the no-longer absolute nature of an event horizon, Hawking dubbed his replacement an apparent horizon.

Of course, this understanding hinges on gravity having wavelike properties. Fortunately for this idea, LIGO has confirmed gravitational waves.

This, like much of what we understand of black holes, is guesswork. From what we've observed, Hawking has been right so far.

[halfeye]

There are equations at Wikipedia for those that like them:

https://en.wikipedia.org/wiki/Hawkin...le_evaporation

The time that the black hole takes to dissipate is:

t e v = 5120 π G 2 M 0 3 ℏ c 4 {\displaystyle t_{\mathrm {ev} }={\frac {5120\pi G^{2}M_{0}^{3}}{\hbar c^{4}}}} {\displaystyle t_{\mathrm {ev} }={\frac {5120\pi G^{2}M_{0}^{3}}{\hbar c^{4}}}}

where M0 is the mass of the black hole.

...

In common units,

P = 3.563 45 × 10 32 [ k g M ] 2 W {\displaystyle P=3.563\,45\times 10^{32}\;\left[{\frac {\mathrm {kg} }{M}}\right]^{2}\;\mathrm {W} } {\displaystyle P=3.563\,45\times 10^{32}\;\left[{\frac {\mathrm {kg} }{M}}\right]^{2}\;\mathrm {W} }

t e v = 8.407 16 × 10 − 17 [ M 0 k g ] 3 s ≈ 2.66 × 10 − 24 [ M 0 k g ] 3 y r {\displaystyle t_{\mathrm {ev} }=8.407\,16\times 10^{-17}\;\left[{\frac {M_{0}}{\mathrm {kg} }}\right]^{3}\;\mathrm {s} \quad \approx \ 2.66\times 10^{-24}\;\left[{\frac {M_{0}}{\mathrm {kg} }}\right]^{3}\;\mathrm {yr} } {\displaystyle t_{\mathrm {ev} }=8.407\,16\times 10^{-17}\;\left[{\frac {M_{0}}{\mathrm {kg} }}\right]^{3}\;\mathrm {s} \quad \approx \ 2.66\times 10^{-24}\;\left[{\frac {M_{0}}{\mathrm {kg} }}\right]^{3}\;\mathrm {yr} }

M 0 = 2.282 71 × 10 5 [ t e v s ] 1 3 k g ≈ 7.2 × 10 7 [ t e v y r ] 1 3 k g {\displaystyle M_{0}=2.282\,71\times 10^{5}\;\left[{\frac {t_{\mathrm {ev} }}{\mathrm {s} }}\right]^{\frac {1}{3}}\;\mathrm {kg} \quad \approx \ 7.2\times 10^{7}\;\left[{\frac {t_{\mathrm {ev} }}{\mathrm {yr} }}\right]^{\frac {1}{3}}\;\mathrm {kg} } {\displaystyle M_{0}=2.282\,71\times 10^{5}\;\left[{\frac {t_{\mathrm {ev} }}{\mathrm {s} }}\right]^{\frac {1}{3}}\;\mathrm {kg} \quad \approx \ 7.2\times 10^{7}\;\left[{\frac {t_{\mathrm {ev} }}{\mathrm {yr} }}\right]^{\frac {1}{3}}\;\mathrm {kg} }

So, for instance, a 1-second-life black hole has a mass of 2.28×105 kg, equivalent to an energy of 2.05×1022 J that could be released by 5×106 megatons of TNT. The initial power is 6.84×1021 W.

Notice that the equations come out looking all wrong in those quotes, because the subscripts etc. are all snarled up.

I'm not sure whether that mass is the same as the one I quoted above, but it's pretty big. Someone can presumably substitute in the other times and get the relevant masses out, if they understand how to work the equations.

[Yora]

So that's 500 gigatons. Or 10,000 Tsar Bombs. 2x10²²J sounds like a lot and covers all the energy consumed in the world for 40 years, but is still only 2% of the energy estimated to be required to heat all the oceans of the world by 1°.

On the cosmic scale of things this is actually pretty low. Even the sun produces 20,000 times as much energy every second. And 10²²J absolutely pales compared to the 10⁴⁴J of a Type 1a supernova.
Unfortunately I can not find out how this compares to the glow of red or white dwarves.

[hamishspence]

Unfortunately I can not find out how this compares to the glow of red or white dwarves.

I'm not sure how much one would need to compensate for the red dwarf emitting much of its radiation as infrared or not.

"Bolometric luminosity" is all electromagnetic radiation, not just visual light.

An example:

https://en.wikipedia.org/wiki/Lacaille_8760

(bolometric luminosity, around 0.07 x sun's, visual luminosity, around 0.02 x sun's)



The lowest figure I've seen for a red dwarf is close to 1/10000 our sun, and the highest on the order of 1/10 our sun.

Given that Proxima's 0.0017 figure is bolometric luminosity, the even smaller 2MASS J0523-1403's 0.000126 figure (around 1/8000 the Sun's), would also appear to be bolometric:

https://www.quora.com/How-large-is-t...est-known-star
https://en.wikipedia.org/wiki/2MASS_J0523-1403

As for white dwarfs - there's one ultra-dim one out there - temperature 2700 C, comparable to dim red dwarfs - PSR J2222-0137B. Given how tiny white dwarf stars are (Earth-sized) this implies a luminosity that is tiny compared to red dwarf stars.

The brightest white dwarfs, however, are on the order of 1/20 the Sun's brightness or even more (there's one super-hot white dwarf star - KPD 0005+5106 with a surface temperature of 200,000 K, but I don't know the exact energy output figure.)

[gomipile]

And 10²²J absolutely pales compared to the 10⁴⁴J of a Type 1a supernova.

The supernova's output is also spread out over weeks of time and a surface area about the same as the planet Earth(the surface area of a white dwarf.)

[factotum]

The supernova's output is also spread out over weeks of time and a surface area about the same as the planet Earth(the surface area of a white dwarf.)

Um, no they're not spread out over weeks? The actual energy release of a supernova explosion happens in seconds. The only reason we see the light curve last for weeks is because of light being emitted from the superhot gases thrown off by the explosion, which expand and slowly cool.

[hamishspence]

Those gases are still producing energy though - via radioactive decay - nickel isotope decaying into lighter elements. The photons produced during the radioactive decay are what heat up the gas of the supernova remnant, and produce the light curve.

[factotum]

Those gases are still producing energy though - via radioactive decay - nickel isotope decaying into lighter elements. The photons produced during the radioactive decay are what heat up the gas of the supernova remnant, and produce the light curve.

Maybe, but as far as I know the 10^44 J figure mentioned earlier is the actual *explosion*, not additional radioactive decay in the Nickel-56 generated in the gas cloud.

(Oh, and note that only applies to Type Ia supernovae--the extended light curve in a Type II supernova is due to ionised hydrogen in the ejecta cloud recombining with electrons to make gaseous hydrogen).

[hamishspence]

(Oh, and note that only applies to Type Ia supernovae--the extended light curve in a Type II supernova is due to ionised hydrogen in the ejecta cloud recombining with electrons to make gaseous hydrogen).

The impression I got was that a Type II Supernova's light curve was dominated by radioactive decay too:

https://en.wikipedia.org/wiki/SN_1987A#Light_curve

Much of the light curve, or graph of luminosity as a function of time, after the explosion of a Type II Supernova such as SN 1987A is provided its energy by radioactive decay. Although the luminous emission consists of optical photons, it is the radioactive power absorbed that keeps the remnant hot enough to radiate light. Without radioactive heat it would quickly dim. The radioactive decay of 56Ni through its daughters 56Co to 56Fe produces gamma-ray photons that are absorbed and dominate the heating and thus the luminosity of the ejecta at intermediate times (several weeks) to late times (several months).[18] Energy for the peak of the light curve of SN1987A was provided by the decay of 56Ni to 56Co (half life 6 days) while energy for the later light curve in particular fit very closely with the 77.3 day half-life of 56Co decaying to 56Fe. Later measurements by space gamma-ray telescopes of the small fraction of the 56Co and 57Co gamma rays that escaped the SN1987A remnant without absorption[19] confirmed earlier predictions that those two radioactive nuclei were the power source.[20]

[Chronos]

We do not know enough to be able to calculate how much of a black hole's radiation occurs in the last second, and we probably won't know enough for a very long time, and anyone who says they do know is going off of a secondhand understanding of the physics which is not relevant in this domain.

You can calculate how much power a black hole will release in any particular kind of particle. But which kinds of particles you have to consider depends on the black hole's temperature, and hence its mass: The higher the hole's temperature, the more massive the particles it can emit. For a stellar-mass black hole, this is easy: The hole is hot enough to produce massless photons and gravitons, and possibly (but probably not) enough for one of the three varieties of neutrinos. So you can add up the power due to photons and gravitons, and possibly the power due to one neutrino variety, integrate it, and find a lifetime for the hole. This lifetime will be accurate enough, since the vast majority of the life will be spent at high mass, where the available set of particles is limited to those.

But in the last moments of its existence, a black hole's temperature would approach the Planck temperature, and so the particles it could emit would approach the Planck mass. And so the list would consist of "all particles of that mass or less". How many unknown particles are out there, that we just haven't discovered yet because our accelerators can't get up to those energies? We have no clue.

Incidentally, there's also another problem here: All of the calculations of Hawking radiation assume that it's a perturbative process, that is, that the hole itself is basically unchanged by each radiation event. That's a great approximation when the mass of the hole is a few times the mass of the Sun and its temperature is a millionth of a kelvin, but it breaks down badly towards the end. And finding the correct assumption to use instead would probably require a working theory of quantum gravity. So that's another difficulty in doing this calculation.

[halfeye]

On the cosmic scale of things this is actually pretty low. Even the sun produces 20,000 times as much energy every second.

If that's right, it's somehow a bit disappointing, I had thought these were supposed to be gamma-ray burst candidates.

We do not know enough to be able to calculate how much of a black hole's radiation occurs in the last second, and we probably won't know enough for a very long time, and anyone who says they do know is going off of a secondhand understanding of the physics which is not relevant in this domain.

You can calculate how much power a black hole will release in any particular kind of particle. But which kinds of particles you have to consider depends on the black hole's temperature, and hence its mass: The higher the hole's temperature, the more massive the particles it can emit. For a stellar-mass black hole, this is easy: The hole is hot enough to produce massless photons and gravitons, and possibly (but probably not) enough for one of the three varieties of neutrinos. So you can add up the power due to photons and gravitons, and possibly the power due to one neutrino variety, integrate it, and find a lifetime for the hole. This lifetime will be accurate enough, since the vast majority of the life will be spent at high mass, where the available set of particles is limited to those.

But in the last moments of its existence, a black hole's temperature would approach the Planck temperature, and so the particles it could emit would approach the Planck mass. And so the list would consist of "all particles of that mass or less". How many unknown particles are out there, that we just haven't discovered yet because our accelerators can't get up to those energies? We have no clue.

Incidentally, there's also another problem here: All of the calculations of Hawking radiation assume that it's a perturbative process, that is, that the hole itself is basically unchanged by each radiation event. That's a great approximation when the mass of the hole is a few times the mass of the Sun and its temperature is a millionth of a kelvin, but it breaks down badly towards the end. And finding the correct assumption to use instead would probably require a working theory of quantum gravity. So that's another difficulty in doing this calculation.

The times are half lives, with the variability that implies. We may not know which particles will come out when, but if we know the mass, we know the overall energy. Presumably some of that energy will be in gravity waves, though my guess would be that the percentage would be very small.

[gomipile]

If that's right, it's somehow a bit disappointing, I had thought these were supposed to be gamma-ray burst candidates.

We're only looking at the very last second, though. That cuts off a lot of the total energy profile.

[Yora]

Hawking radiation is always elementary particles, not full atoms. It's a black hole, nothing that goes inside ever comes out again. The composition is completely irrelevant. The only number that matters is the amount of mass that went into it. What types of matter is irrelevant. In theory you could even add pure energy to a black hole and the effect would be the same as adding an equivalent amount of mass.

I am not even sure what the temperature of a black hole would be. Don't you have to have matter to have a temperature?

[halfeye]

We're only looking at the very last second, though. That cuts off a lot of the total energy profile.

Yeah, however we know that the last second is the most energetic, and if our star is outshining that per second, it's outshining the whole thing, full time. There are some much brighter stars in the galaxy than ours (Eta Carina, the pistol star, Betelguese, UY Scuti, ...).

[Yora]

The sun is one of the smallest stars that can be seen from Earth. I think there are only 7 stars that can be seen without telescopes under ideal conditions that are smaller and dimmer than the sun. An evaporating black hole would have to be very close to see it.

It's not an event you want to accidentally cause in your lab, but out in space it would be nothing but a faint blimp.

[gomipile]

Yeah, however we know that the last second is the most energetic, and if our star is outshining that per second, it's outshining the whole thing, full time. There are some much brighter stars in the galaxy than ours (Eta Carina, the pistol star, Betelguese, UY Scuti, ...).

The evaporating black hole also has the advantage that it is capable of producing photons of arbitrarily high energy the closer it gets to zero Schwarzschild radius.

[Yora]

But that's only for a very short amount of time in a very small volume of space. If you can catch such a photon, it would be a good way to identify the source as an evaporating black hole, though.

[hamishspence]

The sun is one of the smallest stars that can be seen from Earth. I think there are only 7 stars that can be seen without telescopes under ideal conditions that are smaller and dimmer than the sun.

A bit more than that.

Ones listed with an apparent magnitude of less than 6.00 (the limit for suburban skies) but an absolute magnitude of more than the Sun's 4.83 (sometimes this required a little googling as well as using the Wikipedia list:

https://en.wikipedia.org/wiki/List_o...t_bright_stars

Alpha Centauri B (technically you're only seeing the pair, but B would be visible if it was on its own)
Epsilon Eridani
61 Cygni A (61 Cygni B would be at, or slightly below, the unaided eye visibility limit if seen on its own)
Epsilon Indi
Tau Ceti
40 Eridani A
70 Ophiuchi A
Sigma Draconis
Gliese 570
36 Ophiuchi star system (A and B would each be just about visible if on their own)
82 G. Eridani
HR 8832
Xi Boötis A (Xi Boötis B would be below visibility if seen on its own)
Gliese 105 A (others are much too faint to contribute to the system's visible brightness)
HD 4628
107 Piscium
Mu Cassiopeiae A (other is too faint to contribute)
p Eridani A & B (each is close to the unaided eye visibility limit - the combined system will be a little brighter)
Chi Draconis B (A is more luminous than the Sun, B is less luminous but still above magnitude 6)
41 G. Arae A (B is too faint to contribute)
Xi Ursae Majoris B (A is more luminous than the Sun, B is less luminous but still above magnitude 6)
HD_192310
Kappa 1 Ceti
HD 102365/HR 4523/66 G. Centauri A (B is too faint to contribute)
61 Ursae Majoris
HR 4458 A (B is too faint to contribute)
12 Ophiuchi
HR 511
Alpha Mensae
Zeta Herculis B (A is more luminous than the Sun, B is less luminous but still above magnitude 6)
54 Piscium
11 Leonis Minoris A (B is too faint to contribute)
HR 5553 A (B might be contributing a little to this - the whole system is at magnitude 6)
Zeta Reticuli 1
Zeta Reticuli 2 (the Zeta Reticuli system is a wide binary, separable with the eye: 2 is exactly as bright as the Sun, but very slightly less massive)
85 Pegasi A (B is dim enough not to contribute very much)
55 Cancri A (B is too faint to contribute)
HD 69830 (285 G. Puppis)
HD 147513 (62 G. Scorpii) - this one is a bit tricky - absolute magnitude is listed as 4.82 rather than below 4.83 - but luminosity is listed as 0.98x sun's - so I'm going with that.
HD 172051 (86 G. Sagittarii)
58 Eridani
HD 166 (V439 Andromedae - Gliese 5)
Pi 1 Ursae Majoris
Nu 2 Lupi
Psi Serpentis A (B is far too faint to contribute)
HD 38858


So, that's 46 systems with one or more stars in them with a lower luminosity than the Sun, but still above magnitude 6 - counting Zeta Reticuli as two since they are far enough apart that both are visible.


If you restrict it to apparent magnitude 5 however, the number plummets:


Alpha Centauri B
Epsilon Eridani
Epsilon Indi
Tau Ceti
40 Eridani A
70 Ophiuchi A
Sigma Draconis
Xi Boötis A
Xi Ursae Majoris B
Kappa 1 Ceti
HD 102365 A

Only 11 there.

[Leewei]

Hawking radiation is always elementary particles, not full atoms. It's a black hole, nothing that goes inside ever comes out again. The composition is completely irrelevant. The only number that matters is the amount of mass that went into it. What types of matter is irrelevant. In theory you could even add pure energy to a black hole and the effect would be the same as adding an equivalent amount of mass.

I am not even sure what the temperature of a black hole would be. Don't you have to have matter to have a temperature?

How is this take on things reconciled with Conservation of Information?

[Siosilvar]

How is this take on things reconciled with Conservation of Information?

It isn't currently. There are plenty of proposed solutions to the seemingly paradoxical results of black holes, but we don't know.

[Leewei]

It isn't currently. There are plenty of proposed solutions to the seemingly paradoxical results of black holes, but we don't know.

This being the case, I suggest examining the solutions and their implications on black hole evaporation.

Hawking's "apparent horizon" theory indicates that matter will escape as the black hole shrinks.

What else might happen?

[Rockphed]

Can somebody point me to something that explains what "conservation of information" is? Wikipedia points me to an article on intelligent design, which I am pretty sure is not the paradigm espoused by people complaining about black-holes violating "conservation of information".